CHAPTER 6
TWO PORT NETWORKS
1
OBJECTIVES
• To understand about two – port networks
and its functions.
• To understand the different between zparameter, y-parameter, ABCD- parameter
and terminated two port networks.
• To investigate and analysis the behavior of
two – port networks.
2
SUB - TOPICS
 6-1 Z – PARAMETER
 6-2 Y – PARAMETER
 6-3 ABCD – PARAMETER
 6-4 TERMINATED TWO PORT
NETWORKS
3
TWO – PORT NETWORKS
• A pair of terminals through which a current may enter
or leave a network is known as a port.
• Two terminal devices or elements (such as resistors,
capacitors, and inductors) results in one – port
network.
• Most of the circuits we have dealt with so far are two
– terminal or one – port circuits. (Fig. a)
• A two – port network is an electrical network with two
separate ports for input and output.
• It has two terminal pairs acting as access points. The
current entering one terminal of a pair leaves the
other terminal in the pair. (Fig. b)
4
One port or two
terminal circuit
Two port or four
terminal circuit
• It is an electrical
network with two
separate ports for
input and output.
• No independent
sources.
5
6-1 Z – PARAMETER
• Z – parameter also called as impedance parameter and
the units is ohm (Ω)
• The “black box” is replace with Z-parameter is as
shown below.
I
I1
+
V1
-
V1  z11I1  z12I 2
V2  z 21I1  z 22I 2
2
Z11
Z12
Z21
Z22
+
V2
-
V1   z11 z12   I1 
V    z z   I   z 
 2   21 22   2 
where the z terms are called the impedance parameters, or
simply z parameters, and have units of ohms.
 I1 
I 
 2
6
V1
z11 
I1
and
I 2 0
V2
z 21 
I1
I 2 0
z11 = Open-circuit input impedance
z21 = Open-circuit transfer
impedance from port 1 to port 2
V1
z12 
I2
and
I1  0
V2
z 22 
I2
I1  0
z12 = Open-circuit transfer
impedance from port 2 to port 1
z22 = Open-circuit output
impedance
7
Example 1
Find the Z – parameter of the circuit below.
+
V1
I2
I1
+
240Ω
120Ω
_
V2
_
40Ω
8
Solution
When I2 = 0(open circuit port 2). Redraw the circuit.
+
V
1
Ia
I1
240Ω
Ib
120Ω
V1  120 I b .......(1)
+
V2
_
_
40Ω
280
Ib 
I1......(2)
400
sub (1)  (2)
V1
 Z11   84
I1
V2  240 I a .......(3)
120
Ia 
I1.......(4)
400
sub (4)  (3)
V2
 Z 21 
 72
I1
9
When I1 = 0 (open circuit port 1). Redraw the circuit.
Iy
I2
+
V1
+
120Ω
240Ω
_
40Ω
Ix
V2
_
V2  240 I x .......(1)
160
Ix 
I 2 .......( 2)
400
sub (1)  (2)
V2
 Z 22 
 96
I2
V1  120 I y .......(3)
240
Iy 
I 2 .......(4)
400
sub (4)  (3)
V1
 Z12   72
I2
84

In matrix form: Z  
72

72


96
10
Example 2
Find the Z – parameter of the circuit below
I1
+
V1
_
2Ω
10Ω
j4Ω
+
_
I2
+
10I2
-j20Ω
V2
_
11
Solution
i) I2 = 0 (open circuit port 2). Redraw the circuit.
I1
2Ω
j4Ω
I2 = 0
V1  I1 (2  j4)
V1
 (2  j4) 
I1
+
+
 Z11 
V1
_
V2
_
V2  0 (short circuit)
 Z 21  0
12
ii) I1 = 0 (open circuit port 1). Redraw the circuit.
I1 = 0
10Ω
+
V1
_
+
_
I2
V1  10I 2
+
10I2
-j20Ω
V2
_
In matrix form;
0 
(2  j4)
Z  

10
(16
j8)


V1
 Z12 
 10
I2
V2
V2 - 10I 2
I2 

 j20
10
1
 j
2I 2  V2 
 
 20 10 
V2
 Z 22 
 (16 - j8) 
I2
13
6-2 Y - PARAMETER
• Y – parameter also called admittance parameter and
the units is siemens (S).
• The “black box” that we want to replace with the Yparameter is shown below.
I2
I1
+
V1
-
I1  y11V1  y12V2
I 2  y 21V1  y 22V2
Y11
Y12
Y21
Y22
+
V2
-
I1   y11 y12  V1 
V1 
I    y y  V   y V 
 2   21 22   2 
 2
where the y terms are called the admittance parameters, or
simply y parameters, and they have units of Siemens.
14
I1
y11 
V1
and
V2  0
I2
y 21 
V1
V2  0
y11 = Short-circuit input
admittance
y21 = Short-circuit transfer
admittance from port 1 to port 2
y12 
I1
V2
and
V1  0
y 22 
I2
V2
V1  0
y12 = Short-circuit transfer
admittance from port 2 to port 1
y22 = Short-circuit output
admittance
15
Example 3
Find the Y – parameter of the circuit shown
below.
+
V1
_
I1
5Ω
I2
+
20Ω
15Ω
V2
_
16
Solution
i)
+
V1
_
V2 = 0
V1  20 I a .......(1)
5Ω
I1
Ia
20Ω
I2
5
Ia 
I1.......(2)
25
sub (1)  (2)
I1 1
Y11 
 S
V1 4
V1  5 I 2
I2
1
Y21    S
V1
5
17
ii) V1 = 0
I1
5Ω
I2
V2  15 I x .......(3)
+
15Ω
Ix
V2
_
In matrix form;
 1

Y    41

 5
1
5S
4 

15 

5
I 2 .......( 4)
25
sub (3)  (4)
Ix 
 Y22 
I2
4
 S
V2 15
V2  5 I1
I1
1
Y12 
 S
V2
5
18
Example 4
Find the Y – parameters of the circuit
shown.
I1
+
V1
_
2Ω
10Ω
j4Ω
+
_
I2
+
10I2
-j20Ω
V2
_
19
Solution
i) V2 = 0 (short – circuit port 2). Redraw the circuit.
I1
2Ω
10Ω
j4Ω
+
+
V1
_
_
I2
10I2
I0
V1  (2  j4)I1
I1
1
 Y11 

 (0.1 - j0.2) S
V1 2  j4
 Y21 
I2
 0S
V1
20
ii) V1 = 0 (short – circuit port 1). Redraw the circuit.
I1
2Ω
10Ω
j4Ω
+
_
- 10I 2
........(1 )
2  j4
V2
V2 - 10I 2
I2 

- j20
10
I1 
I2
+
10I2
-j20Ω
V2
_
I2
 Y22 
 (0.05  j0.025) S
V2
sub (2)  (1)
Y12 
I1
 (-0.1  j0.075) S
V2
In matrix form;
 1
1 
.......( 2)
2I 2  V2 

0.1  j0.2
10
j20
 Y   



0
 0.1  j0.075 
S

0.05  j0.025 
21
6-3 T (ABCD) PARAMETER
• T – parameter or also ABCD – parameter is a another
set of parameters relates the variables at the input
port to those at the output port.
• T – parameter also called transmission parameters
because this parameter are useful in the analysis of
transmission lines because they express sending – end
variables (V1 and I1) in terms of the receiving – end
variables (V2 and -I2).
• The “black box” that we want to replace with T –
parameter is as shown below.
I2
I1
+
V1
-
A11
B12
C21
D22
+
V2
-
22
V1  AV2  BI 2
I1  CV2  DI 2
V1  A B V2 
V2 
I    C D  I   T  I 
  2
1 
 2
where the T terms are called the transmission parameters,
or simply T or ABCD parameters, and each parameter has
different units.
V
A 1
V2
I1
C
V2
A=open-circuit
voltage ratio
I2 0
I 2 0
C= open-circuit
transfer admittance
(S)
V
B 1
I2
D
I1
I2
V2  0
B= negative shortcircuit transfer
impedance ()
V2  0
D=negative shortcircuit current ratio
23
Example 5
Find the ABCD – parameter of the circuit
shown below.
I1
2Ω
4Ω
+
V1
_
I2
+
10Ω
V2
_
24
Solution
i) I2 = 0,
I1
V2  10 I1
2Ω
+
V1
_
10Ω
+
I1
C 
 0.1S
V2
V2
V1  2 I1  V2
_
6
 V2 
V1  2
  V2  V2
5
 10 
V1
A
 1.2
V2
25
ii) V2 = 0,
I1
2Ω
4Ω
+
V1
10Ω
I1 + I2
_
I2
10
I2  
I1
14
I1
D  
 1.4
I2
V1  2 I1  10I1  I 2 
V1  12 I1  10 I 2
In matrix form;
 1.2
T   
0.1S
6.8
1.4 
 14 
V1  12 
I 2   10 I 2
 10 
V1
B  
 6.8
I2
26
6-4 TERMINATED TWO –
PORT NETWORKS
• In typical application of two port network, the circuit
is driven at port 1 and loaded at port 2.
• Figure below shows the typical terminated 2 port
model.
Zg
Vg
+

I2
I1
+
V1
-
Two – port
network
+
V2
-
ZL
Terminated two-port parameter can be implement to z-parameter,
Y-parameter and ABCD-parameter.
27
• Zg represents the internal impedance of the source and
Vg is the internal voltage of the source and ZL is the
load impedance.
• There are a few characteristics of the terminated twoport network and some of them are;
V1

I1
i)
input impedance, Zi 
ii)
output impedance, Zo 
V2

I2
I2
iii) current gain, A i 
I1
iv) voltage gain, A v 
V2
V1
v) overall voltage gain, A g 
V2
Vg
28
•The derivation of any one of the desired expression
involves the algebraic manipulation of the two – port
equation. The equation are:
1) the two-port parameter equation either Z or Y or ABCD.
For example, Z-parameter, V1  Z11I1  Z12I 2 .......(1)
V2  Z 21I1  Z 22I 2 .......(2)
2) KVL at input,
V1  Vg  I1Zg .......(3)
3) KVL at the output, V  I Z .......(4)
2
2 L
•From these equations, all the characteristic can be obtained.
29
Example 6
For the two-port shown below, obtain the suitable value of
Rs such that maximum power is available at the input
terminal. The Z-parameter of the two-port network is given
as
 Z11
Z
 21
Z12  6


Z 22  4
2
4
With Rs = 5Ω,what would be the value of
Rs
Vs
+

I2
I1
+
V1
-
V2
Vs
Z
+
V2
-
4Ω
30
Solution
Z-parameter equation becomes;
V1  6 I1  2 I 2 .......(1)
V2  4 I1  4 I 2 .......(2)
KVL at the output;
Subs. (3) into (2);
Subs. (4) into (1);
V2  4I 2 .......(3)
I2  
I1
.......( 4)
2
V1  5I1.......(5)
Hence, the input impedance;  Z  V1  5
1
I1
For the circuit to have maximum power,
Rs  Z1  5
31
V
To find 2 at max. power transfer, voltage drop at Z1 is half of
Vs
Vs
Vs
V1  .......(6)
2
From equations (3), (4), (5) & (6),
Overall voltage gain,
V2 1
Ag 

Vs 5
32
Example 7
The ABCD parameter of two – port network shown below
are.
 4
0.1S

20 
2 
The output port is connected to a variable load for a
maximum power transfer. Find RL and the maximum power
transferred.
33
Solution
ABCD parameter equation becomes
(1)
V1 = 4V2 – 20I2
I1 = 0.1V2 – 2I2
(2)
At the input port, V1 = -10I1
(3)
34
(3) Into (1)
-10I1 = 4V2 – 20I2
I1 = -0.4V2 + 2I2
From (5);
ZTH = V2/I2 = 8Ω
(4) Into (2)
-0.4V2 + 2I2 = 0.1V2 – 2I2
(5)
0.5V2 = 4I2
(4)
(6)
But from Figure (b), we know that
V1 = 50 – 10I1 and I2 =0
Sub. these into (1) and (2)
50 – 10I1 = 4V2
(7)
I1 = 0.1V2
(8)
Sub (8) into (7); V2 = 10
Thus, VTH = V2 = 10V
RL for maximum power transfer, RL = ZTH = 8Ω
The maximum power;
P = I2RL = (VTH/2RL)2 x RL = V2TH/4RL = 3.125W
35